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Chapter 5: Q5.3-4E (page 267)

In Exercises \({\bf{3}}\) and \({\bf{4}}\), use the factorization \(A = PD{P^{ - {\bf{1}}}}\) to compute \({A^k}\) where \(k\) represents an arbitrary positive integer.

4. \(\left( {\begin{array}{*{20}{c}}{ - 2}&{12}\\{ - 1}&5\end{array}} \right) = \left( {\begin{array}{*{20}{c}}3&4\\1&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}2&0\\0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 1}&4\\1&{ - 3}\end{array}} \right)\)

Short Answer

Expert verified

The required answer is \({A^k} = \left( {\begin{array}{*{20}{c}}{4 - 3\left( {{2^k}} \right)}&{ - 12 + 3\left( {{2^{k + 2}}} \right)}\\{1 - {2^k}}&{ - 3 + {2^{k + 2}}}\end{array}} \right)\).

Step by step solution

01

Write the Diagonalization Theorem

The Diagonalization Theorem:An \(n \times n\) matrix \(A\) is diagonalizable if and only if \(A\) has \(n\) linearly independent eigenvectors. As \(A = PD{P^{ - 1}}\) which has \(D\) a diagonal matrix if and only if the columns of \(P\) are \(n\) linearly independent eigenvectors of \(A\).

02

Find the inverse of the invertible matrix

Consider the given equation form \(\left( {\begin{array}{*{20}{c}}{ - 2}&{12}\\{ - 1}&5\end{array}} \right) = \left( {\begin{array}{*{20}{c}}3&4\\1&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}2&0\\0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 1}&4\\1&{ - 3}\end{array}} \right)\).

As it is given that \(A = PD{P^{ - 1}}\)than by using the formula for \({n^{th}}\) power we get:

\({A^n} = P{D^n}{P^{ - 1}}\).

Compare the given equation form with \(A = PD{P^{ - 1}}\).

\[\left( {\begin{array}{*{20}{c}}{ - 2}&{12}\\{ - 1}&5\end{array}} \right) = \left( {\begin{array}{*{20}{c}}3&4\\1&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}2&0\\0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 1}&4\\1&{ - 3}\end{array}} \right)\]

Therefore,

\[\begin{array}{c}A = \left( {\begin{array}{*{20}{c}}{ - 2}&{12}\\{ - 1}&5\end{array}} \right)\\P = \left( {\begin{array}{*{20}{c}}3&4\\1&1\end{array}} \right)\\D = \left( {\begin{array}{*{20}{c}}2&0\\0&1\end{array}} \right)\\{P^{ - 1}} = \left( {\begin{array}{*{20}{c}}{ - 1}&4\\1&{ - 3}\end{array}} \right)\end{array}\]

03

Find \({A^k}\)

\[\begin{array}{c}{A^k} = P{D^k}{P^{ - 1}}\\ = \left( {\begin{array}{*{20}{c}}3&4\\1&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{2^k}}&0\\0&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 1}&4\\1&{ - 3}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{3\left( {{2^k}} \right) + 0}&{0 + 4}\\{{2^k} + 0}&{0 + 1}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 1}&{ - 3}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{3\left( {{2^k}} \right)}&4\\{{2^k}}&1\end{array}} \right)\left( {\begin{array}{*{20}{c}}{ - 1}&4\\1&{ - 3}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - 3\left( {{2^k}} \right) + 4}&{3\left( {{2^k}} \right)\left( 4 \right) - 12}\\{ - {2^k} + 1}&{4\left( {{2^k}} \right) - 3}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{4 - 3\left( {{2^k}} \right)}&{ - 12 + 3\left( {{2^{k + 2}}} \right)}\\{1 - {2^k}}&{ - 3 + {2^{k + 2}}}\end{array}} \right)\end{array}\]

Thus, \({A^k} = \left( {\begin{array}{*{20}{c}}{4 - 3\left( {{2^k}} \right)}&{ - 12 + 3\left( {{2^{k + 2}}} \right)}\\{1 - {2^k}}&{ - 3 + {2^{k + 2}}}\end{array}} \right)\).

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Most popular questions from this chapter

Question: Construct a random integer-valued \(4 \times 4\) matrix \(A\), and verify that \(A\) and \({A^T}\) have the same characteristic polynomial (the same eigenvalues with the same multiplicities). Do \(A\) and \({A^T}\) have the same eigenvectors? Make the same analysis of a \(5 \times 5\) matrix. Report the matrices and your conclusions.

Question: Diagonalize the matrices in Exercises \({\bf{7--20}}\), if possible. The eigenvalues for Exercises \({\bf{11--16}}\) are as follows:\(\left( {{\bf{11}}} \right)\lambda {\bf{ = 1,2,3}}\); \(\left( {{\bf{12}}} \right)\lambda {\bf{ = 2,8}}\); \(\left( {{\bf{13}}} \right)\lambda {\bf{ = 5,1}}\); \(\left( {{\bf{14}}} \right)\lambda {\bf{ = 5,4}}\); \(\left( {{\bf{15}}} \right)\lambda {\bf{ = 3,1}}\); \(\left( {{\bf{16}}} \right)\lambda {\bf{ = 2,1}}\). For exercise \({\bf{18}}\), one eigenvalue is \(\lambda {\bf{ = 5}}\) and one eigenvector is \(\left( {{\bf{ - 2,}}\;{\bf{1,}}\;{\bf{2}}} \right)\).

8. \(\left( {\begin{array}{*{20}{c}}{\bf{5}}&{\bf{1}}\\{\bf{0}}&{\bf{5}}\end{array}} \right)\)

For the matrices A in Exercises 1 through 12, find closed formulas for , where t is an arbitrary positive integer. Follow the strategy outlined in Theorem 7.4.2 and illustrated in Example 2. In Exercises 9 though 12, feel free to use technology.

1.A=1203

Let \({\bf{u}}\) be an eigenvector of \(A\) corresponding to an eigenvalue \(\lambda \), and let \(H\) be the line in \({\mathbb{R}^{\bf{n}}}\) through \({\bf{u}}\) and the origin.

  1. Explain why \(H\) is invariant under \(A\) in the sense that \(A{\bf{x}}\) is in \(H\) whenever \({\bf{x}}\) is in \(H\).
  2. Let \(K\) be a one-dimensional subspace of \({\mathbb{R}^{\bf{n}}}\) that is invariant under \(A\). Explain why \(K\) contains an eigenvector of \(A\).

Question: Let \(A = \left( {\begin{array}{*{20}{c}}{ - 6}&{28}&{21}\\4&{ - 15}&{ - 12}\\{ - 8}&a&{25}\end{array}} \right)\). For each value of \(a\) in the set \(\left\{ {32,31.9,31.8,32.1,32.2} \right\}\), compute the characteristic polynomial of \(A\) and the eigenvalues. In each case, create a graph of the characteristic polynomial \(p\left( t \right) = \det \left( {A - tI} \right)\) for \(0 \le t \le 3\). If possible, construct all graphs on one coordinate system. Describe how the graphs reveal the changes in the eigenvalues of \(a\) changes.

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